# Degree of Gauss map equal to half the Euler characteristic and Poincaré-Hopf

The Poincaré-Hopf theorem states that for a smooth compact $m$-manifold $M$ without boundary and a vector field $X\in\operatorname{Vect}(M)$ of $M$ with only isolated zeroes we have the equality $$\sum_{\substack{p\in M\\X(p)=0}}\iota(p,X)=\chi(M)$$ where $\iota(p,X)$ denotes the index of $X$ at $p$ and $\chi(M)$ denotes the Euler characteristic of $M$.

Let $m$ be even and $M\subset\mathbb{R}^{m+1}$ be a $m$-dimensional smooth compact submanifold without boundary and denote by $\nu:M\to S^m$ its Gauss map. How can I deduce from the Poincaré-Hopf theorem that the Brouwer degree of $\nu$ is equal to half the Euler characteristic of $M$ i.e. $$\deg(\nu)=\frac{1}{2}\chi(M)?$$

After many repeated (unsuccessful) tries I was hoping hat someone else might shed some light onto this...

• I recommend that you read Chapter 6 of Milnor's Topology from the differentiable viewpoint and then read up the missing details in his Morse theory.
– t.b.
Aug 25, 2011 at 9:31
• Milnor's "Topology from the differentiable viewpoint" is precisely where I learned about the Poincaré-Hopf theorem. Unfortunately he doesn't address the degree of the Gauss map being half of the Euler characteristic anywhere in this book. Are you saying that in his "Morse theory" there is indeed an exposition of the argument I'm after? Aug 25, 2011 at 11:21
• No I'm not saying that, that is, I don't know. I was convinced that it was in Milnor and I remembered that he referred to Morse theory for some details in the proof of Poincaré-Hopf.
– t.b.
Aug 25, 2011 at 12:32
• This is not true for odd $m$. For example if $M=\mathbb{S}^{1}$ then $\chi(M)=0$, while the degree of the Gauss map equals 1.
– t22
Aug 25, 2011 at 19:17
• Indeed, I'm going to edit my original post to include the assumption of an even dimension. Aug 26, 2011 at 17:28

One way to go about this is to start with your manifold $M \subset \mathbb R^{m+1}$ and consider a height function $f : \mathbb R^{m+1} \to \mathbb R$ (orthogonal projection onto a vector) restricted to $M$. So there is some fixed vector $v \in S^m$ such that $f(x)=\langle x,v\rangle$ for al $x \in \mathbb R^{m+1}$.

Generically, this is a Morse function so its gradient (the orthogonal projection of $v$ to $TM$) is a vector field which is transverse to the $0$-section of $TM$. So Poincare-Hopf tells you how you can compute the Euler characteristic from this.

Now how is that related to the Gauss map? If you were to compute the degree of the Gauss map $\nu : M \to S^m$ by computing its intersection number with $v \in S^m$ you would have a very similar looking sum to compute! But do notice that the orthogonal projection of $v$ to $TM$ can be zero at both $\nu^{-1}(v)$ and $\nu^{-1}(-v)$. When you work out the details this ultimately explains why there's the $1/2$ and why it only works in even dimensions.

I hope that gives you the idea without giving too much away.

• @Sam: I think perhaps you've misunderstood the content of my post. Aug 26, 2011 at 18:52
• @Sam: "Aren't you saying..." My answer is yes. "And that sum..." this question is too vague, I'm not saying anything that vague. "But if we can only..." this seems confused. The point is that solutions to $\nabla (f_{|M}) = 0$ are divided into two sorts -- the ones in $\nu^{-1}(v)$ and the ones in $\nu^{-1}(-v)$. Aug 26, 2011 at 19:27
• This answer definitely helps Aug 26, 2011 at 19:53

Here is a characteristic classes way of doing it.

• One uses the naturally of the Euler class for oriented vector bundles. This means that the Euler class of the induced bundle with induced orientation under a mapping is the pull back of the Euler class, f*(E)

The Euler class of the unit sphere is 1/(2pi)time its volume element. The pull back of the Euler class of the sphere under the Gauss map is the Euler class of the induced bundle - which in this case is the tangent bundle to the surface.

On the other hand the degree of the Gauss map is the integral of the pullback of 1(4pi)volume element of the sphere since the total volume of the sphere is 4pi.

I like Ryan Budney's proof very much. For posterity, I will add my own, different proof. I will use the Jordan-Brouwer separation theorem (see p. 89 of the book by Guillemin & Pollack), the Poincaré-Hopf theorem for manifolds with boundary (which assumes that the vector field is outward pointing at the boundary), and the fact that the Euler characteristic of any odd-dimensional smooth closed manifold vanishes. (In the orientable case, the latter claim follows from Poincaré duality and is exercise 18-9 in Lee's Intro. to Smooth Manifolds, 2nd ed.; the nonorientable case follows from the orientable case, consideration of the orientation double cover, and the fact that the Euler characteristic of a degree $$n<\infty$$ covering space $$E$$ of a compact manifold $$B$$ with boundary satisfies $$\chi(E) = n\cdot \chi(B)$$. Alternatively, see Cor. 3.37 of Algebraic Topology by Hatcher.)

Proof: Let $$M\subset \Bbb{R}^{m+1}$$ be a smooth closed hypersurface. By the Jordan-Brouwer separation theorem, $$M=\partial N$$ is the boundary of some compact, smooth, codimension-$$0$$ manifold $$N$$ with boundary $$\partial N$$. Let $$X$$ be a smooth vector field on $$N$$ which points strictly outward at $$\partial N$$ and has isolated zeros in $$\text{int}(N)$$. By the Poincaré-Hopf theorem, the Euler characteristic $$\chi(N)$$ is equal to the sum of the indices of these zeros. On the other hand, the degree of the Gauss map $$M\to S^m$$ is also equal to the sum of these indices. (To see this, let $$N'$$ be $$N$$ minus the union of small open balls centered at the zeros of $$X$$; the Gauss map $$\frac{X}{|X|}:\partial N'\to S^m$$ admits a smooth extension to all of $$N'$$ and therefore has degree zero, as shown in Guillemin & Pollack.) Hence

$$\chi(N) = \text{degree of the Gauss map }M\to S^m.$$ (1)

Now, in general the following Euler characteristic formula holds for smooth closed manifolds $$N$$ with nonempty boundary:

$$\chi(DN) = 2\chi(N)-\chi(\partial N),$$ (2)

where $$DN$$ is the (compact, boundaryless) double of $$N$$ obtained by pasting two copies of $$N$$ together along their boundaries and smoothing the result. When $$\dim(N)=\dim(DN)$$ is odd, $$\chi(DN)=0$$ as mentioned in the preface.

Therefore, in our case when $$\dim(M) = \dim(\partial N)$$ is even, the left-hand side of (2) vanishes to yield, when combined with (1),

$$\frac{1}{2}\chi(M)= \text{degree of the Gauss map }M\to S^m$$

when $$\dim(M)$$ is even.

This proves the desired result.

Bonus: From (1) and (2) we also obtain the following general formula, valid for arbitrary $$\dim(M) = \dim(\partial N)$$:

$$\chi(N)=\frac{1}{2}[\chi(M)+\chi(DN)]= \text{degree of the Gauss map }M\to S^m$$

where again $$M = \partial N$$ and the existence of $$N$$ is guaranteed by the Jordan-Brouwer separation theorem. Hence if $$\dim(M)=\dim(\partial N)$$ is odd, then $$\chi(M)=0$$ so that

$$\chi(N)=\frac{1}{2}\chi(DN)= \text{degree of the Gauss map }M\to S^m$$

when $$\dim(M)$$ is odd.